text stringlengths 30 4k | source stringlengths 60 201 |
|---|---|
(generically) for a diminishing step-
size (under a Lipschitz condition on ∇gigi)
• Convergence to a “neighborhood” for a constant
stepsize
CONJUGATE DIRECTION METHODS
• Aim to improve convergence rate of steepest
descent, without incurring the overhead of New-
ton’s method
• Analyzed for a quadratic model. They ... | https://ocw.mit.edu/courses/6-252j-nonlinear-programming-spring-2003/5db87e8d64f415730c362e7abd4042a6_6252_slides07.pdf |
choose c(i+1)m so di+1 is Q-conjugate to d0, . . . , di,
(cid:1)
di+1
Qdj = ξi+1
(cid:1)
Qdj +
(cid:4)
(cid:1)
i
(cid:5)(cid:1)
c(i+1)mdm
Qdj = 0.
m=0
d2= ξ2 + c20d0 + c21d1
d1= ξ1 + c10d0
ξ1
ξ2
0
- c10d0
ξ0 = d0
0
d1
d0
CONJUGATE GRADIENT METHOD
• Apply Gram-Schmidt to the vectors ξk = gk =
∇f (xk), k = 0... | https://ocw.mit.edu/courses/6-252j-nonlinear-programming-spring-2003/5db87e8d64f415730c362e7abd4042a6_6252_slides07.pdf |
k is an
inverse Hessian approximation
• Key idea: Successive iterates xk, xk+1 and gra-
dients ∇f (xk), ∇f (xk+1), yield curvature info
qk ≈ ∇2f (xk+1)pk,
pk = xk+1 − xk,
qk = ∇f (xk+1) − ∇f (xk).
(cid:7)−1
(cid:7)(cid:6)
(cid:6)
∇2f (xn) ≈ q0 · · · qn−1 p0 · · · pn−1
• Most popular Quasi-Newton method is a cl... | https://ocw.mit.edu/courses/6-252j-nonlinear-programming-spring-2003/5db87e8d64f415730c362e7abd4042a6_6252_slides07.pdf |
• Coordinate descent.
Applies also to the case
where there are bound
constraints on the vari-
ables.
• Direct search methods. Nelder-Mead method.
PROOF OF CONJUGATE GRADIENT RESULT
• Use induction to show that all gradients gk gen-
erated up to termination are linearly independent.
True for k = 1. Suppose no ter... | https://ocw.mit.edu/courses/6-252j-nonlinear-programming-spring-2003/5db87e8d64f415730c362e7abd4042a6_6252_slides07.pdf |
Physics 8.02T
For now, please sit anywhere, 9 to a table
P01 -
1
Class 1: Outline
Hour 1:
Why Physics?
Why Studio Physics? (& How?)
Vector and Scalar Fields
Hour 2:
Gravitational fields
Electric fields
P01 -
2
Why Physics?
P01 -
3
Why Study Physics?
Understand/appreciate nature
• Lightning
• Soap Films
• Butterfly W... | https://ocw.mit.edu/courses/8-02t-electricity-and-magnetism-spring-2005/5df2d2b1ad70d8dbd050e0e3ac1341cc_presentati_w01d1.pdf |
s?”, Journal of the Learning
Sciences 14(2) 2004.)
Bottom Line: Learn More, Retain More, Do Better
P01 -
10
Overview of TEAL/Studio
Collaborative Learning
Groups of 3, Tables of 9
You teach, you discuss, you learn
In-Class Problem Solving
Desktop Experiments
Teacher-Student Interaction
Visualizations
PRS Questions
P... | https://ocw.mit.edu/courses/8-02t-electricity-and-magnetism-spring-2005/5df2d2b1ad70d8dbd050e0e3ac1341cc_presentati_w01d1.pdf |
Dourmashkin, and Belcher
Supplemental (not required):
Serway & Jewett 6th Edition; Giancoli;
…
Prefer something else? Let me know!
Important: Find something you can read
P01 -
17
Common Questions & Answers
• Dysfunctional Group?
• Must Miss Class?
• Must Miss HW?
• Must Miss Exam?
• Tell Grad TA
• Tell Grad TA
• Tell... | https://ocw.mit.edu/courses/8-02t-electricity-and-magnetism-spring-2005/5df2d2b1ad70d8dbd050e0e3ac1341cc_presentati_w01d1.pdf |
8.02: Electricity and Magnetism
Also new way of thinking…
How do objects interact at a distance?
Fields We will learn about E & M Fields:
how they are created & what they effect
Big Picture Summary:
(cid:119)
∫∫
(cid:71)
(cid:71)
d
E A
⋅
=
Q
in
ε
0
Maxwell
Equations:
S
(cid:71)
(cid:71)
d
B A
⋅
=
0
(cid:119)
∫∫
S
(ci... | https://ocw.mit.edu/courses/8-02t-electricity-and-magnetism-spring-2005/5df2d2b1ad70d8dbd050e0e3ac1341cc_presentati_w01d1.pdf |
04/visualizations/vectorfields/02-particleSource/02-
ParticleSource_320.html)
P01 -
32
Vector Field Examples
Flows With Sinks
(http://ocw.mit.edu/ans7870/8/8.02T/f04/visualizations/vectorfields/01-particleSink/01-
ParticleSink_320.html)
P01 -
33
Vector Field Examples
Circulating Flows
(http://ocw.mit.edu/ans7870/8/8.... | https://ocw.mit.edu/courses/8-02t-electricity-and-magnetism-spring-2005/5df2d2b1ad70d8dbd050e0e3ac1341cc_presentati_w01d1.pdf |
Another Vector Field:
Gravitational Field
P01 -
41
Example Of Vector Field:
Gravitation
(cid:71)
F
g
= −
G
ˆ
r
Gravitational Force:
Mm
2
r
Gravitational Field:
(cid:71)
g
=
(cid:71)
F
g
m
= −
2
/
GMm r
m
ˆ
r
= −
G
M
2
r
ˆ
r
M : Mass of Earth
P01 -
42
Example Of Vector Field:
Gravitation
Gravitational Field:
(cid:71... | https://ocw.mit.edu/courses/8-02t-electricity-and-magnetism-spring-2005/5df2d2b1ad70d8dbd050e0e3ac1341cc_presentati_w01d1.pdf |
electric force between charges q1 and q2 is
(a) repulsive if charges have same signs
(b) attractive if charges have opposite signs
Like charges repel and opposites attract !!
P01 -
47
Coulomb's Law
Coulomb’s Law:
Force by q1 on q2
(cid:71)
=F
12
k
e
q q
1 2
2
r
ˆ
r
ek
=
1
4
πε
0
=
9
8.9875 10 N m /C
×
2
2
ˆ :r unit... | https://ocw.mit.edu/courses/8-02t-electricity-and-magnetism-spring-2005/5df2d2b1ad70d8dbd050e0e3ac1341cc_presentati_w01d1.pdf |
71)
F
23
+
=
In general:
(cid:71)
F
ij
(cid:71)
= ∑F
N
j
i=
1
P01 -
50
Electric Field (~g)
The electric field at a point is the force acting
on a test charge q0 at that point, divided by
the charge q0 :
(cid:71)
F
0q
(cid:71)
E
≡
For a point charge q:
(cid:71)
=E
k
e
q
2
r
ˆ
r
http://ocw.mit.edu/ans7870/8/8.02T/f04/... | https://ocw.mit.edu/courses/8-02t-electricity-and-magnetism-spring-2005/5df2d2b1ad70d8dbd050e0e3ac1341cc_presentati_w01d1.pdf |
Determining n and μ: The Hall Effect
Vx, Ex
+ + + + + + + + + + +
- - - - - - - - -
Ey
I, Jx
r
F
r
+ qv r
= qE
r
× B
Bz
In steady state,
Fy = −evDBz
Fy = −eE y
EY = vDBZ = EH , the Hall Field
Since vD=-Jx/en,
EH = −
1
ne
J xBZ = RH J X BZ
RH = −
1
ne
σ = neμ
Experimental Hall Results on Metals
• V... | https://ocw.mit.edu/courses/3-225-electronic-and-mechanical-properties-of-materials-fall-2007/5e027de7290273530da1882cacc902ca_lecture_2.pdf |
• Microscopic picture
E = E e−iωt
O
Z
e-
dp(t)
dt
= −
p(t)
τ
− eE0 e
−iωt
B=0 in conductor,
r r
r r
and F (E) >> F (B)
− iωp0 = −
try p(t) = p0e−iωt
p0 − eE0
τ
− eE
0
1
τ
p =
0
− iω
ω>>1/τ, p out of phase with E
p0 =
eE0
iω
ω→ ∞, p → 0
ω<<1/τ, p in phase with E
p0 = −eE0τ
What if ωτ>>1?
Whe... | https://ocw.mit.edu/courses/3-225-electronic-and-mechanical-properties-of-materials-fall-2007/5e027de7290273530da1882cacc902ca_lecture_2.pdf |
<<1/τ )
real
Instead of a complex momentum, we can go back to macroscopic
and create a complex J and σ
J (t) = J 0e
− iωτ
J 0 = − nev =
− nep0
m
=
ne2
1
τ
m( − iω )
E0
σ 0
σ =
,σ =
1− iωτ 0
ne2τ
m
Response of e- to AC Electric Fields
•
Low frequency (ω<<1/τ)
– electron has many collisions befor... | https://ocw.mit.edu/courses/3-225-electronic-and-mechanical-properties-of-materials-fall-2007/5e027de7290273530da1882cacc902ca_lecture_2.pdf |
D = ε 0 E + P = εE
r
r
r
B = μ 0 H + μ 0 M = μH
μ= μ rμ 0 ;ε = ε rε 0
r
r
∇ • D = 4πρ
r
∇ • B = 0
r
∇ xE
= −
r
∇ xH =
r
r
r
D = E + 4πP
r
r
r
B = H + 4πM
r
B
1
∂
t
c
∂
r
4π r 1 ∂ D
J +
c ∂ t
c
Waves in Materials
• Non-magnetic material, μ =μ 0
• Polarization non-existent or swamped by free e... | https://ocw.mit.edu/courses/3-225-electronic-and-mechanical-properties-of-materials-fall-2007/5e027de7290273530da1882cacc902ca_lecture_2.pdf |
ω 2
2 ε (ω )
c
ω
k
v =
=
c
ε (ω )
Waves in Materials
• Waves slow down in materials (depends on ε(ω))
• Wavelength decreases (depends on ε(ω))
• Frequency dependence in ε(ω)
ε(ω) = 1+
iσ
ε0ω
= 1+
iσ0
ε0ω(1− iωτ)
ε(ω) = 1+
iω2τ
p
2ω− iωτ
ω2
p =
2
ne
ε0 m
Plasma Frequency
For ωτ>>>1, ε(ω) goes to 1... | https://ocw.mit.edu/courses/3-225-electronic-and-mechanical-properties-of-materials-fall-2007/5e027de7290273530da1882cacc902ca_lecture_2.pdf |
>>1
ω<ωp, ε is negative, k=ki, wave reflected
ω>ωp, ε is positive, k=kr, wave propagates
(
εω
) = 1
−
ω2
p
2
ω
R
ω
p
ω
Success and Failure of Free e- Picture
• Success
– Metal conductivity
– Hall effect valence=1
– Skin Depth
– Wiedmann-Franz law
• Examples of Failure
– Insulators, Semiconductors
– Hal... | https://ocw.mit.edu/courses/3-225-electronic-and-mechanical-properties-of-materials-fall-2007/5e027de7290273530da1882cacc902ca_lecture_2.pdf |
.1
2.3
1.5
0.52
0.80
1.13
1.0
2.38
0.88
0.5
0.64
0.38
0.09
0.18
k s T
(watt-ohm K2)
2.22 x 108
2.12
2.23
2.42
2.20
2.31
2.32
2.36
2.14
2.90
2.61
2.28
2.49
2.14
2.58
2.75
2.48
2.64
3.53
2.57
373K
k
(watt cm-K)
0.73
k s T
(watt-ohm K2)
2.43 x 108
3.82
4.17
3.1
1.7
1.5
0.54
0.73
1.1
1.0
2.30
0.80
0.45
0.60
0.35
0.08
0.17
... | https://ocw.mit.edu/courses/3-225-electronic-and-mechanical-properties-of-materials-fall-2007/5e027de7290273530da1882cacc902ca_lecture_2.pdf |
Class 5 – Project Choice Discussion
• Everyone should have a copy of the Project System Title
Table Handout
• Write your (readable) name on the line near the bottom
• Identify where in the list your “Title” (as shortened in many
cases by the faculty) appears
•
• Add the following late additions (anyone not appearing)
... | https://ocw.mit.edu/courses/esd-342-advanced-system-architecture-spring-2006/5e18ea1129b7d261f0046332fe676306_lec5.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
6.334 Power Electronics
Spring 2007
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/6-334-power-electronics-spring-2007/5e2f970d6c7c0ad911c62677e1b9fb82_ch6.pdf |
6.895 Essential Coding Theory
September 15, 2004
Lecturer: Madhu Sudan
Scribe: Adi Akavia
Lecture 3
Today’s Plan
• Converse coding theorem
• Shannon vs. Hamming theories
• Goals for the rest of the course
• Tools we use in this course
Shannon’s Converse Theorem
We complete our exposition of Shannon’s theory ... | https://ocw.mit.edu/courses/6-895-essential-coding-theory-fall-2004/5e3930912bd217554cd58c0573ca7a39_lect03.pdf |
message m. On the other hand, the average number of received
words that decode to m is 2n−k , since D : {0, 1} ≥ {0, 1}k , for k = Rn. Now, since 2H(p)n >> 2n−k ,
then each corrupted word is mapped back to its original with only a very small probability.
n
More formally, fix some encoding and decoding mappings E, D. ... | https://ocw.mit.edu/courses/6-895-essential-coding-theory-fall-2004/5e3930912bd217554cd58c0573ca7a39_lect03.pdf |
with very small probability exp(−n).
Therefore,
Pr[correct decoding] � exp(−n) + Pr[correct decoding|� has more than p�n non-zero entries]
= exp(−n) +
�
�
� /�B(p� n,n) m
Pr[message m, error �, Im,� = 1]
3-1
As the choices of message, error and decoding algorithm are independent, and Pr[m] = 2−k, Pr... | https://ocw.mit.edu/courses/6-895-essential-coding-theory-fall-2004/5e3930912bd217554cd58c0573ca7a39_lect03.pdf |
as the error has some finite variance, then we might as well think of the channel as having some
finite capacity (even when the alphabet in infinite as in R!).
Shannon also considers more general probability distributions on the error of the channel. In particu
lar, Shannon considers Markovian error model. Markovian mo... | https://ocw.mit.edu/courses/6-895-essential-coding-theory-fall-2004/5e3930912bd217554cd58c0573ca7a39_lect03.pdf |
Theory
The main target arising from Shannon’s theory is to explicitly find efficient encoding and decoding
function. In particular, for Binary Symmetric channel, BSCp, can we come up with polynomial-time
encoding and decoding functions? Or, better yet, linear-time functions?
3-2
Shannon tells us that for any rate ... | https://ocw.mit.edu/courses/6-895-essential-coding-theory-fall-2004/5e3930912bd217554cd58c0573ca7a39_lect03.pdf |
the size of the code is
|Code| =
�k
�
�
�
�
. We’d like to maximize k.
• d is the minimum distance of code d = minx⊥=y �Hamming (x, y). We’d like to maximize d.
• q is the alphabet size |�|. It is not clear whether want q to be minimized of maximized. Nonetheless,
we general assume that we want to minimize q, as ... | https://ocw.mit.edu/courses/6-895-essential-coding-theory-fall-2004/5e3930912bd217554cd58c0573ca7a39_lect03.pdf |
Definition 2 (Field). A Field (F, +, ·, 0, 1) is a set F with addition and multiplication operations +, ·
(respectively) and special elements 0, 1 such that:
• addition forms a commutative group on the elements of F,
• multiplication forms a commutative group on the elements of F \ {0}, and
• there is a distribution... | https://ocw.mit.edu/courses/6-895-essential-coding-theory-fall-2004/5e3930912bd217554cd58c0573ca7a39_lect03.pdf |
is vector addition (i.e., (v1, ..., vn) + (u1, ..., un) = (v1 + u1, ..., vn + un)), and · is a scalar
product (i.e., �(v1, ..., vn) = (�v1, ..., �vn)).
Linear Codes
We’d like the codes we construct to have nice properties such as succinct representation, efficient encod
ing, and efficient decoding. Linear codes (as defi... | https://ocw.mit.edu/courses/6-895-essential-coding-theory-fall-2004/5e3930912bd217554cd58c0573ca7a39_lect03.pdf |
0 implies �1, .., �k = 0).
An alternate specification of Linear subspace is by its null space:
C � { y | ∈y, v≈ = 0�v ≤ C}
(where the inner product ∈x, y≈ =
�
linear, and dim(C �) + dim(C) = n.
xiyi for any x, y ≤ F). For linear codes C, the null space C � is also
Empirically, there seem to be no harm in restricti... | https://ocw.mit.edu/courses/6-895-essential-coding-theory-fall-2004/5e3930912bd217554cd58c0573ca7a39_lect03.pdf |
MIT 6.035
MIT 6 035
Introduction to Shift-Reduce Parsing
Martin Rinard
Laboratory for Computer Science
Massachusetts Institute of Technology
Orientation
• Specify Syntax Using
Context-Free Grammar
• Nonterminals
• Terminals
• Productions
• Given a grammar, Parser
Generator produces a
Generator produces a ... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
ser Exam p lep
Expr → Expr Op Expr
Expr → ( Expr)
Expr → - Expr
Expr → num
Op → +
p
Op → -
Op → *
Expr
Op
Expr
num
*
num
+
num
)
Expr
(
Op
Expr
T
F
F
I
H
S
Shift-Reduce Parser Examplep
Expr → ExprOp Expr
Expr→ (Expr)
Expr→ - Expr
Expr→ num
Op→ +
p
Op→ -
Op→ *
)
Expr
E
(
Expr
Op
Expr
num
*
num
+
num... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
Potential Conflicts
• Reduce/Reduce Conflict
t
k
• Top of the stack may match RHS of multiple
h
f
T
t
productions
Which production to use in the reduction?
• Which production to use in the reduction?
•
RHS
l
t
h
t
f
i
l
• Shift/Reduce Conflict
Stack may match RHS of production
• Stack may match RHS of produ... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
T
T
F
I
H
S
S
num
-
num
Shift/Reduce/Reduce Conflict
Expr → ExprOp Expr
Expr → Expr- Expr
Expr→ (Expr)
Expr→ Expr -
Expr→ num
Op→ +
Op→ -
Op →
Op → *
*
p
What Happens if
What Happens if
Choose
Reduce
num
Expr
T
T
F
I
H
S
S
Expr
num
-
Shift/Reduce/Reduce Conflict
Expr → ExprOp Expr
Expr → E... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
*
*
p
What Happens if
What Happens if
Choose
Reduce
Expr
Op
Expr
Expr
T
P
E
E
C
C
A
A
num
-
num
Conflicts
What Happens if
What Happens if
Choose
Shift
num
num
-
Expr
T
F
F
I
H
S
Expr → ExprOp Expr
Expr → Expr- Expr
Expr→ (Expr)
Expr→ Expr -
Expr→ num
Op→ +
Op→ -
Op →
Op → *
*
p
Confli... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
Expr
Expr→ (Expr)
Expr→ Expr -
Expr→ num
Op→ +
Op→ -
Op →
Op → *
*
p
Shift/Reduce/Reduce Conflict
/
This Shift/Reduce Conflict
Reflects Ambiguity in
Grammar
-
Expr
Expr → ExprOp Expr
Expr → Expr- Expr
Expr→ (Expr)
Expr→ Expr -
Expr→ num
Op→ +
Op→ -
Op →
Op → *
*
p
num
num
Shift/Reduce/Reduce Conf... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
Integrating Finite State Control
• Actions
P h S b l
d S O S k
• Push Symbols and States Onto Stacks
• Reduce According to a Given Production
••
Accept
Accept
t t
t
t
• Selected action is a function of
Current input
Current input
•
•
s
symbol
• Current state of finite state control
Each action specifies next ... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
4
reduce (2)
reduce (3)
hift t
$
error
accept
error
error
reduce (2)
reduce (3)
Goto
X
goto s1
goto s3
State
s0
s1
s2
s3
3
s4
s5
(
shift to s2
error
shift to s2
error
reduce (2)
reduce (3)
• Shift to sn
• Push input token into the symbol stack
• Push sninto state stack
• Advance to next input symbol
Parser Tab... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
ACTION
)
error
error
shift to s5
shift to s4
4
reduce (2)
reduce (3)
hift t
$
error
accept
error
error
reduce (2)
reduce (3)
Goto
X
goto s1
goto s3
State
s0
s1
s2
s3
3
s4
s5
(
shift to s2
error
shift to s2
error
reduce (2)
reduce (3)
• Accept
• Stop parsing and report success
Parse Table In Action
State
s0
s1
s... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
s5
(
shift to s2
error
shift to s2
error
reduce (2)
reduce (3)
ACTION
)
error
error
shift to s5
shift to s4
4
reduce (2)
reduce (3)
hift t
$
error
accept
error
error
reduce (2)
reduce (3)
Goto
X
goto s1
goto s3
State Stack Symbol Stack
Input
Grammar
s2
s2
s0
(
())$
S → X$ (1)
X→ (X ) (2)
X→ ( ) (3)
X→ (
) (3)
P... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
) (3)
X→ (
) (3)
Parse Table In Action
State
s0
s1
s2
s3
3
s4
s5
(
shift to s2
error
shift to s2
error
reduce (2)
reduce (3)
ACTION
)
error
error
shift to s5
shift to s4
4
reduce (2)
reduce (3)
hift t
$
error
accept
error
error
reduce (2)
reduce (3)
Goto
X
goto s1
goto s3
State Stack Symbol Stack
Input
Grammar
... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
goto s1
goto s3
State Stack Symbol Stack
Input
Grammar
s5
s2
s2
s2
s0
)
(
(
(
)$
S → X$ (1)
X→ (X ) (2)
X→ ( ) (3)
X→ (
) (3)
Step One: Pop Stacks
p
p
State
s0
s1
s2
s3
3
s4
s5
(
shift to s2
error
shift to s2
error
reduce (2)
reduce (3)
ACTION
)
error
error
shift to s5
shift to s4
4
reduce (2)
reduce (3)
hift t... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
error
shift to s2
error
reduce (2)
reduce (3)
ACTION
)
error
error
shift to s5
shift to s4
4
reduce (2)
reduce (3)
hift t
$
error
accept
error
error
reduce (2)
reduce (3)
Goto
X
goto s1
goto s3
State Stack Symbol Stack
Input
Grammar
s2
s2
s0
(
)$
S → X$ (1)
X→ (X ) (2)
X→ ( ) (3)
X→ (
) (3)
Step Two: Push Nonte... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
) (2)
X→ ( ) (3)
X→ (
) (3)
Step Three: Use Goto, Push New State
p
,
State
s0
s1
s2
s3
3
s4
s5
(
shift to s2
error
shift to s2
error
reduce (2)
reduce (3)
ACTION
)
error
error
shift to s5
shift to s4
4
reduce (2)
reduce (3)
hift t
$
error
accept
error
error
reduce (2)
reduce (3)
Goto
X
goto s1
goto s3
State Sta... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
error
error
reduce (2)
reduce (3)
Goto
X
goto s1
goto s3
State Stack Symbol Stack
Input
Grammar
s4
s3
s2
s2
s0
)
X
X
(
$
S → X$ (1)
X→ (X ) (2)
X→ ( ) (3)
X→ (
) (3)
Parse Table In Action
State
s0
s1
s2
s3
3
s4
s5
(
shift to s2
error
shift to s2
error
reduce (2)
reduce (3)
ACTION
)
error
error
shift to s5
shift... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
s1
s2
s3
3
s4
s5
(
shift to s2
error
shift to s2
error
reduce (2)
reduce (3)
ACTION
)
error
error
shift to s5
shift to s4
4
reduce (2)
reduce (3)
hift t
$
error
accept
error
error
reduce (2)
reduce (3)
Goto
X
goto s1
goto s3
State Stack Symbol Stack
Input
Grammar
$
S → X$ (1)
X→ (X ) (2)
X→ ( ) (3)
X→ (
) (3)
s0
... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
(3)
s0
X
Step Three: Use Goto, Push New State
p
,
State
s0
s1
s2
3
s3
s4
s5
(
shift to s2
error
shift to s2
error
reduce (2)
reduce (3)
ACTION
)
error
error
shift to s5
shift to s4
4
reduce (2)
reduce (3)
hift t
$
error
accept
error
error
reduce (2)
reduce (3)
Goto
X
goto s1
goto s3
State Stack Symbol Stack
Inp... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
1
goto s3
State Stack Symbol Stack
Input
Grammar
s1
s1
s0
S
$
S → X$ (1)
X→ (X ) (2)
X→ ( ) (3)
X→ (
) (3)
Key Concepts
t
•
•
t t
parser a
St k Fi i
t
c
controlling parser actions
• Pushdown automaton for parsing
• Stack, Finite state control
l
• Parse actions: shift, reduce, accept
Parse table for
Parse ... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
6.801/6.866: Machine Vision, Lecture 12
Professor Berthold Horn, Ryan Sander, Tadayuki Yoshitake
MIT Department of Electrical Engineering and Computer Science
Fall 2020
These lecture summaries are designed to be a review of the lecture. Though I do my best to include all main topics from the
lecture, the lectures ... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
“rules” of patents:
– No equations are included in the patent (no longer true)
– No greyscale images - only black and white
– Arcane grammar is used for legal purposes - “comprises”, “apparatus”, “method”, etc.
– References of other patents are often included - sometimes these are added by the patent examiner, rath... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
08,109). Each of these sections will be discussed in further detail below. Before we get into the specifics of the patent again,
it is important to point out the importance of edge detection for higher-level machine vision tasks, such as:
• Attitude (pose estimation) of an object
• Object recognition
• Determining t... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
where the second derivative of brightness crosses
zero, a.k.a. where r(ru(x)) = r u(x) = 0. Notice that the location of this zero is given by the same location as the inflection
point of u(x) and the maximum of ru(x):
2
Gradient2 of “Soft” Unit Step Function, ru(x)
0.1
5 · 10−2
)
x
(
u
2
r
0
−5 · 10−2
−0.1
−6
... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
2 −1 0 1
−1 1
3. Ex = 2 −1 1
1
3
Where for molecule 2, the best point for estimating derivatives lies directly in the center pixel, and for molecules 1 and 3, the
best point for estimating derivatives lies halfway between the two pixels.
How do we analyze the efficacy of this approach?
1. Taylor Ser... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
nd-order derivatives, we just apply another
derivative operator, which is equivalent to convolution of another derivative estimator “molecule”:
∂2
∂x2
(·) =
∂(·)
∂
∂x ∂x
⇐⇒ −1 1 ⊗ −1
1
1 =
1
2
1 −2
1
1
For deriving the sign here and understanding why we have symmetry, remember that co... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
2
((1 ∗ −1) + (−2 ∗ 0) + (1 ∗ 1)) =
1
2
(−1 + 0 + 1) = 0
Where we note that = 1 due to the pixel spacing. This is equivalent to f 00(x) = 0.
4
• f (x) = 1:
f (x) = 1
f 0(x) = 0
f 00(x) = 0
Applying the 2nd derivative estimator above to this function:
1 −2 1 ⊗
1
f (−1) = 1 f (0) = 1 f (1) = 1 ... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
derivative estimator operators, the weights of the “stencils”/computational molecules should add up to zero. Now
that we have looked at some of these operators and modes of analysis in one dimension, let us now look at 2 dimensions.
1.2.3 Mixed Partial Derivatives in 2D
First, it is important to look at the linear, ... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
5 degrees counterclockwise in the 2D plane: x = x cos 45+y cos 45 =
rotated coordinate system, Ex x = Exy .
√
2
2
x+
2
2
√
0
0
0
0
0
• Intuition for convolution: If convolution is a new concept for you, check out reference [2] here. Visually, convolution
is equivalent to “flipping and sliding” one operator across... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
tools:
• Taylor Series
• Test functions
• Fourier analysis
Intuitively, we know that neither of these estimators will be optimal, because neither of these estimators are rotationally-
symmetric. Let us combine these intelligently to achieve rotational symmetry. Adding four times the first one with one times
the sec... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
Another technique leveraged in this patent was Non-Maximum Suppression (NMS). Idea: Apply edge detector estimator
operator everywhere - we will get a small response in most places, so what if we just threshold? This is an instance of early
decision-making, because once we take out these points, they are no longer con... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
the patent focuses on the interpolation technique used for sub-pixel gradient plotting for peak finding.
To find an optimal interpolation technique, we can plot s vs. s , where s = s|2s|b , where b ∈ N is a parameter that determines
0
the relationship between s and s0 .
0
In addition to cubic interpolation, we can als... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
We will discuss this in greater detail next lecture.
Multiscale is quite important in edge detection, because we can have edges at different scales. To draw contrasting exam-
ples, we could have an image such that:
• We have very sharp edges that transition over ≈ only 1 pixel
• We have blurry edges that transition o... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
from defocusing, but this
defocusing compensation helps.
1.3.2 Addressing Quantization of Gradient Directions
Here is another possible extension not included in the edge detection patent.
Recall that because spaces occurs in two sizes (pixel spacing and
2 pixel spacing), we need to sample in two ways, which
can l... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
a change of coordinates from cartesian
to polar:
(Ex, Ey ) → (E0, Eθ)
Idea: Rotate a coordinate system to make estimates using test angles iteratively. Note that we can simply compute these with
square roots and arc tangents, but these can be prohibitively computationally-expensive:
q
E0 = Ex
2
2 + Ey
Eθ = a... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
π π
,
8
4
1
2i
1
1
Note that this reduces computation to 2 additions per iteration. Angle we turn through becomes successively smaller:
r
cos θi =
1 + → R =
1
22i
Y
i
cos θi =
r
1 +
1
22i
Y
i
≈ 1.16 (precomputed)
1.4 References
1. Finite Differences, https://en.wikipedia.org/wiki/Finite difference
__________... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
6.801/6.866: Machine Vision, Lecture 20
Professor Berthold Horn, Ryan Sander, Tadayuki Yoshitake
MIT Department of Electrical Engineering and Computer Science
Fall 2020
These lecture summaries are designed to be a review of the lecture. Though I do my best to include all main topics from the
lecture, the lectures ... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
solids, with 4, 6, 8, 12, and
20 faces. Each of the tessellations from the platonic solids results in equal area projections on the sphere, but the division is
somewhat coarse.
For greater granularity, we can look at the 14 Archimedean solids. This allows for having multiple polygons in each
polyhedra (e.g. squares... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
cally, hyperboloids of one sheet, that lead to
ambiguity in the solution space for relative orientation problems.
Why are Critical Surfaces important? Critical surfaces can negatively impact the performance of relative orientation
systems, and understanding their geometry can enable us to avoid using strategies that... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
5. Cone: c2 = a2 + b2
2
x
2
Figure 5: 3D geometric depiction of a cone, another member of the quadric family and a special case of the hyperboloid of one
sheet.
6. Elliptic Paraboloid: = +
z
c
2
x
a2
2
y
b2
Note that this quadric surface has a linear, rather than quadratic dependence, on z.
Figure 6: 3D geome... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
-view geometry, is known and
calibrated (usually this results in finding a calibration matrix K). To achieve high-performing binocular stereo systems, we need
to find the relative orientation between the two cameras. A few other notes on this:
• Calibration is typically for binocular stereo using baseline calibration.... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
Places that pass through epipolar lines are called
epipolar planes, as given below:
Figure 9: Epipolar planes are planes that pass through epipolar lines.
Next, in the image, we intersect the image plane with our set of epipolar planes, and we look at the intersections of these
image planes (which become lines). Th... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
/scenes we image), we cannot
get absolute length of the baseline b, and therefore, we treat the baseline as a unit vector. Treating the baseline as a
unit vector results in one less DOF and 5 DOF for the system overall (since we now only have 2 DOF for the unit vector
baseline).
1.3.2 How Many Correspondences Do We... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
and actually takes the constraints into account. In practice,
vertical disparity is tuned out first using an iterative sequence of 5 moves.
In practice, we would like to use more correspondences for accuracy. With more correspondences, we no longer try to find
exact matches, but instead minimize a sum of squared error... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
product of these three (hopefully near coplanar) vectors. This is a feasible approach, but
also have a high noise gain/variance.
This leads us to an important question: What, specifically, are we trying to minimize? Recall that we are matching cor-
respondences in the image, not in the 3D world. Therefore, the volume ... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
3
dot products:
0
1. ·(r
l × rr
): This yields the following:
Lefthand side :
Righthand side :
Combining : γ||r 0 × r
0
(b + βrr) · (r
l × rr
2 = [b r 0
l rr]
r||2
l
0
0
(αr
l + γ(rl × rr)) · (rl × rr) = γ||rl × rr
0
0
0
) = b · r
l × rr
2
2||
0
l rr
+ 0 = [b r
]
Intuitively, this says that the erro... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
+ γ(r
l × rr
(b + βrr
) · ((rr × (rl × rr
0
0
)
= (b × rr) · (r
||
α||r
l × rr
l × rr
0
0
2
l × rr
)) · ((rr × (r
||
l × rr
)) = α||r
2
0
0
)) = (b × rr) · (r
)
l × rr
2
2
Taking this dot product with our equation allows us to find α.
With this, we have now isolated all three of our desired parameters.... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
equation, which we do not (yet) have the closed-form solutions to, we cannot solve for this problem in
closed-form. However, we can still solve for it numerically. We can also look at solving this through a weighted least-squares
approach below.
1.3.4 Solving Using Weighted Least Squares
Because our distance d can ... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
of the weights as being the conversion factor from 3D to 2D error, and
therefore, they can roughly be thought of as w = Z , where f is the focal length and Z is the 3D depth.
f
Now that we have expressed a closed-form solution to this problem, we are ready to build off of the last two lectures and
apply our unit quat... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
At first glance, it appears we have 8 unknowns, with 5 constraints. But we can add additional constraints to the system to make
the number of constraints equal the number of DOF:
1. Unit quaternions: ||q||2 = q · q = 1
o
o
o
2. Unit baseline: ||b||2 = b · b = 1
o
o o
3. Orthogonality of q and d: q · d = 0
o
o
o... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
Assume/fix one of the two unknowns d
or q. This results in a linear objective and a linear problem to solve, which, as we know, can be solved with least squares
approaches, giving us a closed-form solution:
o
o
• Option 1: Assume q is known and fix it −→ solve for d
o
o
• Option 2: Assume d is known and fix it −→ sol... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
correspondences, we
don’t have enough constraints to find solutions.
• Gef¨arliche Fla¨achen, also known as “dangerous or critical surfaces”: There exist surfaces that make the relative
orientation problem difficult to solve due to the additional ambiguity that these surfaces exhibit. One example: Plane
flying over a v... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
)(dx + ey + f ) = adx2 + aexy + af x + bdxy + bey2 + bf y + cdx + cey + cf = 0
We can see that this equation is indeed second order with respect to its spatial coordinates, and therefore belongs to the
family of quadric surfaces and critical surfaces.
Figure 15: The intersection of two planes is another type of crit... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
to optimize.
• e(θ) is the residual function of the objective evaluated with the current set of parameters.
Note the λI, or regularization term, in Levenberg-Marquadt. If you’re familiar with ridge regression, LM is effectively ridge
regression/regression with L2 regularization for nonlinear optimization problems. Of... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
Dynamics of a Single Particle (Review) (continued)
1
2.003J/1.053J Dynamics and Control I, Spring 2007
Professor Thomas Peacock
2/12/2007
Lecture 2
Work-Energy Principle
Dynamics of a Single Particle (Review) (continued)
Reading: Williams 4–1, 4–2, 4–3 (Momentum Principles) - End of lecture last
time
Work-Energy... | https://ocw.mit.edu/courses/2-003j-dynamics-and-control-i-spring-2007/5eaa7b1b85eff45f6a92b5a6d6052665_lec02.pdf |
2 −
1
2
2
mv 1
1
2
=
=
Define Kinetic Energy T = 1
⇒ W12 = T2 − T1 = (Work done is the change in the kinetic energy)
2 m|v|2
Consider the case:
F =
−∂V
∂r
For example,
⇒ W12 =
r2
�
r
1
F · dr =
Thus, in this special case:
F = (
∂V
∂x
r2 −∂V
∂r
�
r
1
ˆı −
∂V
∂y
jˆ)
· dr = V1 − V2(Potential Ener... | https://ocw.mit.edu/courses/2-003j-dynamics-and-control-i-spring-2007/5eaa7b1b85eff45f6a92b5a6d6052665_lec02.pdf |
, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology.
Downloaded on [DD Month YYYY].
Dynamics of a Single Particle (Review) (continued)
3
Figure 2: Free body diagram. The only force acting on the mass is the force
due to gravity mg. Figure by MIT OCW.
Figure 3: Mass... | https://ocw.mit.edu/courses/2-003j-dynamics-and-control-i-spring-2007/5eaa7b1b85eff45f6a92b5a6d6052665_lec02.pdf |
1. Scalar Equation
2. Need not account for forces that do no work
Example 1: Vehicle Falling on a Curve
A small vehicle is released from rest at the top of a circular path. Determine
the angle θ0 to the position where the vehicle leaves the path and becomes a
projectile.
(Neglect friction and treat vehicle as a p... | https://ocw.mit.edu/courses/2-003j-dynamics-and-control-i-spring-2007/5eaa7b1b85eff45f6a92b5a6d6052665_lec02.pdf |
sin θˆı − cos θjˆ) = mR θ¨ eˆt − mRθ˙2 eˆn
F = −mgjˆ+ N eˆn = (N − mg cos θ)ˆen + mg cos θeˆt
�
From Newton II:
F = p˙
�
N − mg cos θ = −mR θ˙2
Normal
mg sin θ = mR θ ¨
Tangential
(1)
(2)
We have:
−mg cos θ + N = −mR θ˙2
Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053... | https://ocw.mit.edu/courses/2-003j-dynamics-and-control-i-spring-2007/5eaa7b1b85eff45f6a92b5a6d6052665_lec02.pdf |
(1 − cos θ)
Putting into (1) and letting N = 0, Rθ˙2 = R 2
g (1 − cos θ) = g cos θ0.
R
2(1 − cos θ0) = cos θ0
cos θ0 =
θ0 = 48.2
2
3
◦
Note: This solution is independent of R and m.
Linear momentum principle used to solve problem.
Next time, use angular momentum principle to solve a problem.
Cite as: Thomas Peac... | https://ocw.mit.edu/courses/2-003j-dynamics-and-control-i-spring-2007/5eaa7b1b85eff45f6a92b5a6d6052665_lec02.pdf |
18.657: Mathematics of Machine Learning
Lecturer: Philippe Rigollet
Scribe: Ali Makhdoumi
Lecture
6
Sep. 28, 2015
5. LEARNING WITH A GENERAL LOSS FUNCTION
In the previous lectures we have focused on binary losses for the classification problem and
developed VC theory for it. In particular, the risk for a classification f... | https://ocw.mit.edu/courses/18-657-mathematics-of-machine-learning-fall-2015/5ebb42429b252cbd2f711cd03e01b97f_MIT18_657F15_L6.pdf |
those materials and we will
refer to the techniques we have used in classification when needed.
[
−
[
−
∈
5.1 Empirical Risk Minimization
5.1.1 Notations
Loss function: In binary classification the loss function was 1I(h(X) = Y ). Here, we
replace this loss function by ℓ(Y, f (X)) which we assume is symmetric, where f
,
... | https://ocw.mit.edu/courses/18-657-mathematics-of-machine-learning-fall-2015/5ebb42429b252cbd2f711cd03e01b97f_MIT18_657F15_L6.pdf |
compare its performance with the following oracle:
ˆ
¯f
∈
argmin R(f ).
f ∈F
Note that this is an oracle as in order to find it one need to have access to PXY and then
optimize R(f ) (we only observe the data Dn). Since f is the minimizer of the empirical
ˆ
risk minimizer, we have that Rn(f )
≤
ˆ
ˆR(f )
ˆ
Rn(f ) + Rn(f ... | https://ocw.mit.edu/courses/18-657-mathematics-of-machine-learning-fall-2015/5ebb42429b252cbd2f711cd03e01b97f_MIT18_657F15_L6.pdf |
ˆ
sup Rn(f )
|
f ∈F
−
R(f )
| ≤
E
I
ˆ
sup Rn(f )
|
"
∈
f
F
−
R(f )
|#
+
r
log (1/delta)
2n
, w.p. 1
.
δ
−
As a result we only need to bound the expectation IE[supf ∈F |
ˆ
Rn(f )
].
R(f )
|
−
2
6
5.1.2 Symmetrization and Rademacher Complexity
Similar to the binary loss case we first use symmetrization technique and then... | https://ocw.mit.edu/courses/18-657-mathematics-of-machine-learning-fall-2015/5ebb42429b252cbd2f711cd03e01b97f_MIT18_657F15_L6.pdf |
i=1
X
1
n
"
1
n
"|
n
i=1
X
n
ℓ(Yi, f (X
i))
1
− n
ℓ(Yi, f (X
i))
i=1
X
ℓ(Yi, f (Xi))
1
n
−
1
n
−
n
i=1
X
n
i=1
X
n
i=1
X
n
ℓ(Y ′
i , f (X ′
i)) Dn
|
ℓ(Yi , f (X ′
′
i)) Dn
|
# |#
# |#
′
ℓ(Y , f (X ′
i
i))
Dn
| |
##
i=1
X
ℓ(Y ′
i , f (X ′
i))
|#
|#
i=1
X
n
1
sup
f ∈F | n
(c)
≤
2IE
"
i=1
X
1
sup
f ∈F n
|
2 sup IE
Dn
≤
"
... | https://ocw.mit.edu/courses/18-657-mathematics-of-machine-learning-fall-2015/5ebb42429b252cbd2f711cd03e01b97f_MIT18_657F15_L6.pdf |
i, f (xi))
|#
.
Therefore, we have
IE
1
sup
f ∈F n
|
"
n
i=1
X
ℓ(Yi, f (Xi))
−
IE[ℓ(Yi, f (Xi))]
|# ≤
n(ℓ
2
R
◦ F
)
and we only require to bound the Rademacher complexity.
5.1.3 Finite Class of functions
Suppose that the class of functions
F
is finite. We have the following bound.
3
Theorem: Assume that
F
is finite and ... | https://ocw.mit.edu/courses/18-657-mathematics-of-machine-learning-fall-2015/5ebb42429b252cbd2f711cd03e01b97f_MIT18_657F15_L6.pdf |
(this is not exactly what we used but the XOR argument of the previous lecture
risk of f
allows us to show that the cardinality of this set is the same as the cardinality of the set
that interests us). In this lecture, this set might be uncountable. Therefore, we need to
introduce a metric on this set so that we can tr... | https://ocw.mit.edu/courses/18-657-mathematics-of-machine-learning-fall-2015/5ebb42429b252cbd2f711cd03e01b97f_MIT18_657F15_L6.pdf |
. However, for
applying the results in order to bound empirical risk minimization, we take xi to be (xi, yi)
and
. We define the empirical l1 distance as
to be ℓ
F
F
◦ F
dx
1(f, g) =
1
n
n
i
=1
X
f (x )
i
|
.
g(xi)
|
−
Theorem: If 0
f
≤
≤
1 for all f
∈ F
, then for any x = (x1, . . . , xn), we have
ˆRx
n(
F
)
inf
ε≥0
≤
... | https://ocw.mit.edu/courses/18-657-mathematics-of-machine-learning-fall-2015/5ebb42429b252cbd2f711cd03e01b97f_MIT18_657F15_L6.pdf |
|#
Since the previous bound holds for any ε, we can take the infimum over all ε
0 to obtain
≥
x
Rn(
F
)
≤
inf ε +
ε≥0
(cid:8)
2 log(2N (
F
n
r
, dx
1 , ε))
.
(cid:9)
The previous bound clearly establishes a trade-off because as ε decreases N (
creases.
F
, dx
1 , ε) in-
5.2.2 Computing Covering Numbers
As a warm-up, we w... | https://ocw.mit.edu/courses/18-657-mathematics-of-machine-learning-fall-2015/5ebb42429b252cbd2f711cd03e01b97f_MIT18_657F15_L6.pdf |
∪
⊆
≤
where vol(B2) denotes the volume of the unit Euclidean ball in d dimensions. It yields,
ε
( )dvol(B2)
2
|
V
|
(1 + )dvol(B2) ,
ε
2
V
|
| ≤ (cid:0)
1 + ε d
2
d
ε
2
(cid:1)
=
2
ε
(cid:18)
d
+ 1
(cid:19)
≤
(cid:18)
3 d
ε
(cid:19)
.
(cid:0)
(cid:1)
6
For any p
1, define
≥
and for p =
, define
∞
dx
p(f, g) =
f (xi)
|
−... | https://ocw.mit.edu/courses/18-657-mathematics-of-machine-learning-fall-2015/5ebb42429b252cbd2f711cd03e01b97f_MIT18_657F15_L6.pdf |
n
1
zi
|
p) p
|
max
i
zi
|
,
|
≤
i=1
X
p, ε) and N (f, d∞, ε)
which leads to B(f, dx
q <
1
B(f, dx
. Using H¨older’s inequality with r = q
∞, ε)
⊆
p
≤
≤
∞
≥
1 we obtain
p ≥
N (f, dp, ε). Now suppose that
1
p
z
i
|
p
|
!
1
n
n
i=1
X
1
−n p
≤
n
(1
−
1
1
)
r p
n
1
!
i 1
=
X
zi
|
pr
|
!
i=1
X
1
pr
=
1
q
.
zi
|
q
|
!
1
n
n
... | https://ocw.mit.edu/courses/18-657-mathematics-of-machine-learning-fall-2015/5ebb42429b252cbd2f711cd03e01b97f_MIT18_657F15_L6.pdf |
log n
n
).
7
MIT OpenCourseWare
http://ocw.mit.edu
18.657 Mathematics of Machine Learning
Fall 2015
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/18-657-mathematics-of-machine-learning-fall-2015/5ebb42429b252cbd2f711cd03e01b97f_MIT18_657F15_L6.pdf |
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